Bài 1: Tập hợp Q các số hữu tỉ

\(\left(\dfrac{2}{5}\right)^9\cdot5^9\)

\(=\left(\dfrac{2}{5}\cdot5\right)^9\)

\(=\left(\dfrac{10}{5}\right)^9\)

\(=2^9\)

\(=512\)

\(\left(\dfrac{1}{5}\right)^5\cdot5^5\)

\(=\left(\dfrac{1}{5}\cdot5\right)^5\)

\(=\left(\dfrac{5}{5}\right)^5\)

\(=1^5\)

\(=1\)

`@` `\text {Ans}`

`\downarrow`

`(1/5)^5*5^5`

`= (1/5*5)^5`

`= (5/5)^5`

`= 1^5 = 1`

=>1/2*2^(x+4)-2^x=2^10(2^3-1)

=>2^(x+3)-2^x=2^10(2^3-1)

=>2^x=2^10

=>x=10

\(M=\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{2001}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(M=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{4002}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(M=\left(\dfrac{1}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right)\)

\(M=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)

\(M=1:5\)

\(M=\dfrac{1}{5}\)

\(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)

=1/5

\(=\left(-2-\dfrac{7}{12}\right):\dfrac{15}{7}-\dfrac{1}{18}:\dfrac{15}{7}+\dfrac{20}{9}:\dfrac{15}{7}\)

\(=\dfrac{7}{15}\left(-2-\dfrac{7}{12}-\dfrac{1}{18}+\dfrac{20}{9}\right)\)

\(=\dfrac{7}{15}\cdot\dfrac{-72-21-2+80}{36}=\dfrac{7}{15}\cdot\dfrac{-15}{36}=\dfrac{-7}{36}\)

\(\left(-2\dfrac{7}{12}\right):2\dfrac{1}{7}-\dfrac{1}{18}:2\dfrac{1}{7}+2\dfrac{2}{9}:2\dfrac{1}{7}\)

\(=\left(-2\dfrac{7}{12}-\dfrac{1}{18}+2\dfrac{2}{9}\right):2\dfrac{1}{7}\)

\(=\left(\dfrac{-31}{12}-\dfrac{1}{18}+\dfrac{20}{9}\right):2\dfrac{1}{7}\)

\(=-\dfrac{5}{12}:\dfrac{15}{7}\)

\(=-\dfrac{5}{12}\cdot\dfrac{7}{15}\)

\(=-\dfrac{7}{36}\)

\(\dfrac{1}{3}< x< \dfrac{1}{2}\)

=> \(\dfrac{1}{3}=1:3=0,333...\) ; \(\dfrac{1}{2}=1:2=0,5\)

Mà \(\dfrac{3}{8}=0,375\)

Nên \(\dfrac{1}{3}< \dfrac{3}{8}< \dfrac{1}{2}\)

P/s: đây là ý kiến riêng tớ mình nhé cou, chúc cou hok tốt :3

Rất nhiều giá trị được em ha: 7/20; 19/50; 39/100; 81/200; 9/19; 2/5,...

=3/2*7/3+4/3*1/2

=21/6+4/6=25/6

\(\dfrac{11}{2}.\dfrac{21}{3}+\dfrac{11}{3}.\dfrac{1}{2}\) 

\(=\dfrac{77}{2}+\dfrac{11}{6}=\dfrac{121}{3}\)

ủa z là đáp án nào đúng trùi

cả lò nhà mik ơi mik nói lun đây là tính hợp lí ạ

=13/30-1-5/6+2/5+17/14+1/2-5/7+5/67

=-1+17/14-3/14+5/67

=-1+1+5/67

=5/67

Bài 11: 

Ta có: \(x=\dfrac{-101}{a+7}\) nguyên khi \(-101⋮a+7\)

Vậy: \(a+7\inƯ\left(101\right)\)

\(Ư\left(101\right)=\left\{101;1;-101;-1\right\}\)

\(a+7\in\left\{101;1;-101;-1\right\}\)

\(\Rightarrow a\in\left\{94;-108;-6;-8\right\}\)

Vậy x sẽ nguyên khi \(a\in\left\{94;-108l-6;-8\right\}\)

Bài 12:

Ta có: \(t=\dfrac{3x+8}{x-5}=\dfrac{3x+15-7}{x-5}=\dfrac{3\left(x+5\right)-7}{x-5}=3+\dfrac{7}{x-5}\)

t nguyên khi \(\dfrac{7}{x+5}\) nguyên tức là \(x-5\inƯ\left(7\right)\) 

\(Ư\left(7\right)=\left\{-7;7;-1;1\right\}\)

\(\Rightarrow x-5\in\left\{-7;7;-1;1\right\}\)

\(\Rightarrow x\in\left\{12;-2;4;6\right\}\)

Vậy t sẽ nguyên khi \(x\in\left\{12;-2;4;6\right\}\)